LinkedList Struct Typedef in C

Question

I am try to build a basic linked list in C. I seem to get an error for this code:

typedef struct
{
    char letter;
    int number;
    list_t *next;
}list_t         


        

Answer 1

When you are using list_t *next in your code, the compiler doesn't know what to do with list_t, as you haven't declared it yet. Try this:

typedef struct list {
    char letter;
    int number;
    struct list *next;
} list;

As H2CO3 pointed out in the comments, using _t as an identifier suffix is not a great idea, so don't use list_t.

Answer 2

why did you make it hard on yourself just set openGame and ruzeLopez as pointers and you wont have to use the & (this is the usual way to use linked lists , just don't forget to use -> to access members)

try this code instead :

#include <stdlib.h>
#include <malloc.h>

typedef struct list
{
    char letter;
    int number;
    struct list *next;
}list;

main(void)
{
   char letters[] = "ABCDEFGH"; //what were the braces for ?
   list *openGame, *ruyLopez;
   openGame = ruyLopez = NULL;
   openGame = malloc(sizeof(list));
   ruyLopez = malloc(sizeof(list));

   openGame->letter = letters[4];
   openGame->number = 4;
   openGame->next = ruyLopez;

   ruyLopez->letter = letters[5];
   ruyLopez->number = 5;
   ruyLopez->next = NULL;
}

Answer 3

Here is a working example without the risk of memory leaks from using malloc to create your structures.

#include <stdlib.h>

typedef struct _list
{
    char letter;
    int number;
    struct _list *next;
} list_type;

int main(void)
{
   char letters[] = "ABCDEFGH"; //what were the braces for ?
   list_type openGame, ruyLopez;
   openGame.letter = letters[4];
   openGame.number = 4;
   openGame.next = &ruyLopez;

   ruyLopez.letter = letters[5];
   ruyLopez.number = 5;
   ruyLopez.next = NULL;
}